Stiffness Matrix Method

1. Introduction — Why Matrix Methods?

The Slope-Deflection Method and Moment Distribution Method work well for structures with a small number of members and joints. For a two-span beam or a single-bay portal frame, writing and solving a handful of equations by hand is perfectly manageable. But real structures — multi-storey building frames, bridge decks, offshore platforms — have hundreds or thousands of members and joints. Solving hundreds of simultaneous equations by hand is not feasible.

Matrix methods solve this problem by expressing the structural analysis in a standardised matrix format that can be implemented directly in a computer program. Every member is described by the same type of matrix (the element stiffness matrix). These matrices are assembled into a single system (the global stiffness matrix) using the same algorithm regardless of the structure’s size or complexity. Solving the matrix equation [K]{D} = {F} for the unknown displacements {D} takes the computer a fraction of a second, even for large structures.

The two main matrix methods are:

• Stiffness Method (Displacement Method): Unknowns are displacements. System size = number of free degrees of freedom. Preferred for structures with many members (most practical cases).
• Flexibility Method (Force Method): Unknowns are redundant forces. System size = degree of static indeterminacy. Preferred for structures with high degrees of freedom but low indeterminacy (rare in practice).

All modern structural analysis software — STAAD.Pro, SAP2000, ETABS, ANSYS, Abaqus — uses the stiffness method internally. Understanding how it works gives engineers the ability to check software output critically, understand the limitations of the models, and diagnose errors. This understanding is also directly tested in GATE CE.

2. Degrees of Freedom and Kinematic Indeterminacy

A degree of freedom (DOF) is an independent displacement component at a joint — a translation in x, a translation in y, or a rotation about z (in 2D analysis). The total number of degrees of freedom in a structure determines the size of the stiffness matrix.

Relationship between DSI and DKI: These are different quantities. DSI (static indeterminacy) determines the number of compatibility equations needed in the force method. DKI (kinematic indeterminacy) determines the number of displacement unknowns in the stiffness method. For the same structure, DSI and DKI can be very different. However: DSI + DKI = constant for a given structure, related by the member count and connectivity.

3. Stiffness Method vs Flexibility Method

4. Local Stiffness Matrix — Beam Element

A beam element has two end nodes (near end i and far end j), each with 2 DOFs: vertical displacement (v) and rotation (θ). The element therefore has 4 DOFs total. The local stiffness matrix [k] is a 4×4 matrix that relates the four end forces {Q} (shear forces and moments at each end) to the four end displacements {d} (vertical displacements and rotations at each end).

         v_i        θ_i       v_j       θ_j
    ┌                                         ┐
v_i │  12EI/L³   6EI/L²  -12EI/L³   6EI/L²  │
θ_i │  6EI/L²    4EI/L   -6EI/L²    2EI/L   │
v_j │ -12EI/L³  -6EI/L²   12EI/L³  -6EI/L²  │
θ_j │  6EI/L²    2EI/L   -6EI/L²    4EI/L   │
    └                                         ┘
    

    ┌  12    6L   -12    6L  ┐
    │  6L    4L²  -6L    2L² │
    │ -12   -6L    12   -6L  │
    └  6L    2L²  -6L    4L² ┘
    

Key observations about the beam stiffness matrix:

• It is symmetric: k_ij = k_ji (Maxwell’s reciprocal theorem).
• Its determinant is zero (singular): the matrix cannot be inverted directly — this is because the element can undergo rigid body motion (translation and rotation) without any deformation. The matrix becomes invertible only after boundary conditions are applied.
• The diagonal entries are positive: applying a unit displacement in any DOF direction requires a positive force/moment in the same direction.
• The columns sum to zero: a unit displacement with no external constraint produces no net force on the element (equilibrium).

5. Assembly of Global Stiffness Matrix

The global stiffness matrix [K] is assembled by superimposing the contributions of all element stiffness matrices. Each element stiffness matrix entry is added to the global matrix at the position corresponding to the element’s global DOF numbers.

Assembly procedure (Direct Stiffness Method):

1. Number all DOFs globally: Assign a unique number to every DOF in the structure (free and restrained). This creates the global DOF numbering.
2. Identify element DOF connectivity: For each element, list the global DOF numbers corresponding to its local DOFs.
3. Initialise [K] as a zero matrix of size (total DOF) × (total DOF).
4. For each element: Add the element’s local stiffness matrix entries to the global matrix at the positions indicated by the element’s DOF connectivity. If element local DOF 1 corresponds to global DOF 3, and local DOF 2 corresponds to global DOF 5, then k₁₂ is added to K₃₅.
5. Repeat for all elements. Entries from multiple elements sharing a DOF (internal joints) are added together.

6. Applying Boundary Conditions

After assembly, the global stiffness matrix [K] relates all DOFs (free and restrained) to all forces. Before solving, boundary conditions must be applied to account for the known zero displacements at supports.

    ┌ K_ff  K_fr ┐ ┌ D_f ┐   ┌ F_f ┐
    │            │ │     │ = │     │
    └ K_rf  K_rr ┘ └ D_r ┘   └ F_r ┘
    

Alternative (elimination) method: Delete the rows and columns of [K] corresponding to the restrained DOFs (where displacement = 0). The remaining reduced matrix [K_ff] is the one to invert. This is the simplest approach for hand calculations.

7. Complete Solution Procedure

1. Discretise the structure: Identify nodes (joints, supports, points of load application, points of geometric change) and elements (members between nodes).
2. Number DOFs globally: Assign global DOF numbers to all nodal displacement components.
3. Formulate element stiffness matrices: Compute [k] for each element using the appropriate formula (beam, truss, or frame element).
4. Assemble global stiffness matrix [K]: Superimpose all element stiffness matrices using the DOF connectivity.
5. Formulate load vector {F}: Assemble the nodal load vector from all applied forces and moments. For loads along member lengths (UDL, point loads not at nodes), convert to equivalent nodal loads using fixed-end forces.
6. Apply boundary conditions: Identify restrained DOFs and reduce [K] to [K_ff].
7. Solve for displacements: {D_f} = [K_ff]⁻¹{F_f}.
8. Recover element end forces: For each element: {Q} = [k]{d} − {Q₀}, where {d} = element end displacements (from global solution) and {Q₀} = fixed-end forces from member loads.
9. Find support reactions: {F_r} = [K_rf]{D_f}.
10. Draw SFD and BMD: Using the recovered end forces and known member loads.

8. Properties of the Stiffness Matrix

The properties of the stiffness matrix are frequently tested in GATE CE as conceptual MCQ questions. Memorise all of them.

9. Worked Example 1 — Propped Cantilever by Stiffness Method

Problem: A propped cantilever AB is fixed at A and supported by a roller at B. Span L. Carries a UDL of w per unit length. Using the stiffness matrix method, find the end moments and support reactions. EI constant.

    [k] = (EI/L³) ×
    DOF:  1      2      3      4
    ┌  12    6L   -12    6L  ┐  DOF 1
    │  6L    4L²  -6L    2L² │  DOF 2
    │ -12   -6L    12   -6L  │  DOF 3
    └  6L    2L²  -6L    4L² ┘  DOF 4
    

10. Worked Example 2 — Two-Span Continuous Beam

Problem: A two-span continuous beam ABC: pin at A, roller at B (intermediate), roller at C. Span AB = L, span BC = L. UDL of w on span AB only. EI constant. Find end moments using the stiffness method.

         θ_A    θ_B    θ_C
    ┌  4EI/L  2EI/L    0   ┐ θ_A
    │  2EI/L  8EI/L  2EI/L │ θ_B  (4+4 at B from both elements)
    └    0    2EI/L  4EI/L ┘ θ_C
    

    (EI/L) × [4  2  0]
             [2  8  2]
             [0  2  4]
    

11. Stiffness Matrix for Truss Element

A truss element resists only axial force. Each node has 2 DOFs (u and v — horizontal and vertical translations). The element stiffness matrix in local coordinates (along the member axis) is 2×2:

    [k_local] = (AE/L) × [ 1  -1]
                         [-1   1]
    

    [k_global] = [T]ᵀ[k_local][T]
    

    [k_global] = (AE/L) ×
    ┌  c²    cs   -c²   -cs ┐
    │  cs    s²   -cs   -s² │
    │ -c²   -cs    c²    cs │
    └ -cs   -s²    cs    s² ┘
    

12. Stiffness Matrix for Frame Element

A frame element resists both axial force AND bending. Each node has 3 DOFs: u (horizontal), v (vertical), θ (rotation). The element has 6 DOFs total. The stiffness matrix is 6×6, combining the axial stiffness (AE/L terms) with the bending stiffness (EI/L terms).

         u_i    v_i     θ_i      u_j    v_j     θ_j
    ┌  AE/L     0       0      -AE/L    0       0    ┐
    │   0    12EI/L³  6EI/L²    0   -12EI/L³  6EI/L²│
    │   0    6EI/L²   4EI/L    0    -6EI/L²  2EI/L  │
    │ -AE/L   0       0       AE/L    0       0      │
    │   0   -12EI/L³ -6EI/L²   0    12EI/L³ -6EI/L² │
    └   0    6EI/L²   2EI/L    0    -6EI/L²  4EI/L  ┘
    

13. Common Mistakes Students Make

• Confusing stiffness coefficient K_ij with flexibility coefficient f_ij: K_ij is the force at DOF i due to a unit displacement at DOF j. F_ij is the displacement at DOF i due to a unit force at DOF j. These are inverses of each other. Mixing them up — for example, writing K_ij as a displacement — is a fundamental conceptual error. Always remember: stiffness = force per unit displacement; flexibility = displacement per unit force.
• Forgetting to convert member loads to equivalent nodal loads: The stiffness method works with nodal forces only. Any load applied along a member (UDL, intermediate point load) must be converted to equivalent nodal forces and moments before assembly. Applying member loads directly to the stiffness equation without this conversion gives wrong results. The equivalent nodal loads are the negatives of the fixed-end forces.
• Not applying boundary conditions before inverting [K]: The global stiffness matrix before applying boundary conditions is singular (determinant = 0) and cannot be inverted. Trying to solve [K]{D} = {F} directly without first reducing to [K_ff] will always fail. Boundary conditions must be applied first by eliminating restrained DOF rows and columns.
• Wrong DOF numbering leading to assembly errors: The entire assembly depends on correctly identifying which global DOF corresponds to each local element DOF. An error in DOF numbering causes the wrong entries to be superimposed, producing a completely wrong global stiffness matrix. Always draw the structure with numbered DOFs before starting assembly.
• Ignoring axial deformation in frames: For most practical frame problems in GATE CE, axial deformation of members is neglected (the axial stiffness AE/L is assumed infinite — members are inextensible). This reduces the DOFs significantly — horizontal displacements are equal for all nodes in the same column, and the stiffness matrix is smaller. Failing to apply this assumption where expected leads to unnecessary complexity; failing to apply it where the problem requires it (e.g., when members are explicitly stated to be axially rigid) leads to wrong results.

14. Frequently Asked Questions

Why is the stiffness matrix singular before boundary conditions are applied?

The global stiffness matrix before applying boundary conditions is singular because the structure as a whole can undergo rigid body motion — translation and rotation — without any member deforming. When the structure moves as a rigid body, no strain energy is stored, so the stiffness matrix produces zero force for these displacement modes. Mathematically, the rigid body displacements are the null space of [K], making its determinant zero. Applying boundary conditions (fixing supports) eliminates the rigid body modes by constraining specific DOFs to zero, removing the null space and making the reduced stiffness matrix [K_ff] positive definite and invertible.

What is the difference between the stiffness matrix method and the Moment Distribution Method?

Both are displacement methods — they express end moments in terms of joint rotations. The Moment Distribution Method is an iterative, hand-calculation procedure that reaches the solution by successive approximation (distributing and carrying over moments until convergence). The stiffness matrix method is a direct, exact method that sets up the complete system of equations and solves them simultaneously in one step. MDM is faster for small structures by hand and gives physical insight into how moments redistribute. The stiffness matrix method is implemented in computer programs and handles any structure of arbitrary complexity. The end results (joint rotations, end moments) are identical for both methods applied to the same structure.

What is the bandwidth of a stiffness matrix and why does it matter?

The bandwidth of a stiffness matrix is the maximum distance from the diagonal of any non-zero entry — it measures how “banded” (clustered near the diagonal) the matrix is. A smaller bandwidth means the non-zero entries are concentrated close to the diagonal, with large blocks of zeros elsewhere. This matters enormously for computational efficiency: solving a banded system of equations requires far fewer operations than solving a full (dense) system. For a stiffness matrix of size n×n with bandwidth b, the solution requires O(nb²) operations instead of O(n³). In large structures, minimising bandwidth by optimal DOF numbering (node renumbering algorithms like Cuthill-McKee) can reduce computation time by orders of magnitude.

How does the stiffness method handle support settlement?

Support settlement is handled elegantly in the stiffness method. If a support settles by a known amount δ, the corresponding DOF is not zero — it equals δ. In the partitioned equation [K]{D} = {F}, the settled DOF is a restrained DOF with a known non-zero value. The term [K_fr]{D_r} in the equation [K_ff]{D_f} = {F_f} − [K_fr]{D_r} becomes non-zero, adding an additional load effect to the right-hand side. This automatically accounts for the settlement-induced forces in the structure. This is one of the advantages of the stiffness method over hand calculation methods like MDM, where settlement requires separate treatment using the settlement FEM formulae (6EIδ/L²).